想了2天了，hibernate的更新问题，请高手指点迷津，紧急求助（为何我的更新操作会变成插入操作？？？？？？）

Question

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Answer 1

这段是批量更新的sample:
调用update方法就可以了，下面是update里使用到的queryActionByFind和updateAction方法，这样更新一个和多个都可以了，加分吧朋友！！！
public void update(){
String[] columnNames=EmployeeDetailTable.getColumnNames();
Object[] values={"null","a00X","null","null","null","null","null","null","null","null"};
Hashtable msgDetails=new Hashtable();
for(int i=0;i<columnNames.length;i++){
msgDetails.put(columnNames[i],values[i]);
}
Iterator it=EmployeeDetailTable.queryActionByFind(
"from EmployeeDetail e where e.employeeid='a000'");
EmployeeDetailTable.updateAction(it,msgDetails);
}public static Iterator queryActionByFind(String HQL){
Session session=HibernateUtility.currentSession();
Iterator i=null;
try {
i = session.find(HQL).iterator();
} catch (HibernateException e) {
e.printStackTrace();
}
return i;
}public static void updateAction(Iterator it,Hashtable msgDetails){
EmployeeDetail employeeDetail=null;
Session session=HibernateUtility.currentSession();
String[] columnNames=getColumnNames();
while(it.hasNext()){
employeeDetail=(EmployeeDetail)it.next();
for(int i=0;i<columnNames.length;i++){
//Hashtable中不支持null,只能用"null"代替
if(!msgDetails.get(columnNames[i]).toString().equals("null")){
switch(i){
case 0:
employeeDetail.setSysid((Integer)msgDetails.get(columnNames[i]));
break;
case 1:
employeeDetail.setEmployeeid((String)msgDetails.get(columnNames[i]));
break;
case 2:
employeeDetail.setName((String)msgDetails.get(columnNames[i]));
break;
case 3:
employeeDetail.setGender((String)msgDetails.get(columnNames[i]));
break;
case 4:
employeeDetail.setAge((Integer)msgDetails.get(columnNames[i]));
break;
case 5:
employeeDetail.setAddress((String)msgDetails.get(columnNames[i]));
break;
case 6:
employeeDetail.setZipcode((String)msgDetails.get(columnNames[i]));
break;
case 7:
employeeDetail.setMobile((String)msgDetails.get(columnNames[i]));
break;
case 8:
employeeDetail.setUnitphone((String)msgDetails.get(columnNames[i]));
break;
case 9:
employeeDetail.setEmail((String)msgDetails.get(columnNames[i]));
break;
default:
}
}
}
}
try {
session.flush();
session.evict(employeeDetail);
HibernateUtility.closeSession();
} catch (HibernateException e) {
e.printStackTrace();
}
}

Answer 2

补充一下上面的EmployeeDetailTable.getColumnNames()方法是从自己写的xml文件里得到employeedetail表的列名，省得自己一个个去写了，xml原型如下，当然你也可以用final static String[]定死
<?xml version="1.0" encoding="UTF-8"?>
<Table name="EmployeeDetail">
<Column name="sysid" type="INT" isNull="NOT NULL"/>
<Column name="employeeid" type="CHAR(20)" isNull="NOT NULL"/>
<Column name="name" type="CHAR(20)" isNull="NOT NULL"/>
<Column name="gender" type="CHAR(1)" isNull="NOT NULL">
<ValueOptional>M</ValueOptional>
<ValueOptional>F</ValueOptional>
</Column>
<Column name="age" type="SMALLINT" isNull="NOT NULL"/>
<Column name="address" type="CHAR(100)" isNull="NULL"/>
<Column name="zipcode" type="CHAR(20)" isNull="NULL"/>
<Column name="mobile" type="CHAR(20)" isNull="NULL"/>
<Column name="unitphone" type="CHAR(20)" isNull="NULL"/>
<Column name="email" type="CHAR(50)" isNull="NULL"/>
</Table>